A load of $270 \mathrm{kN}$ is applied on a short concrete column $250 \mathrm{mm} \times 250 \mathrm{mm}$. The column reinforced with 8 bars of $16 \mathrm{mm}$ diameter. If the modulus of elasticity for steel is 18 times that of concrete, find the stress in concrete and steel. If the steel in concrete shall not exceed $4 \mathrm{N} / \mathrm{mm}^{2}$, find the area of steel required so that the Column may support a load of $400 \mathrm{kN}$.

Case 1: When the column carries a load of 270KN

Area of steel $As=\frac{\pi}{4}*16^2*8=1608.48mm^2$

Area of concrete=$A_c=250*250-1608.48=60891.52 mm^2$

Let the stress in steel and concrete be $P_s$ and $P_c$

Strain in steel=strain in concrete

$\frac{P_s}{E_s}=\frac{P_c}{E_c}$

$P_s=\frac{E_s}{E_c}*P_c$

i.e $P_s=18P_c$ ( since, $E_s = 18 E_c$ ) --- (1)

Now, load on steel+load on concrete=load on column

$P_sA_s+P_cA_c=P$ ( since, Load = Stress * Area )

$18P_c*1608.5+P_c*60891.52 = 270000N$ (from - (1))

$P_c=3 N/mm^2$ ---------------------------- (2)

$P_s=18*3=54 N/mm^2$ --------------from (1) & (2)

Case 2: When the column carries a load of 400KN

Let the area of steel be as $mm^2$

Area of concrete=$A_c=250*250-A_s mm^2$

$A_c=62500-A_s mm^2$

Stress in concrete $P_c= 4 N/mm^2$

Stress in steel $P_s=18*P-c=18*4=72 N/mm^2$ ------ (from 1)

Load on steel+load on concrete=load on column

$P_sA_s+P_cA_c=P$

$72*A_s+4(62500-A_s)=400000N$

$A_s=2205.9 mm^2$